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Every union of connected components is a module, and so is every union of anticonnected components

Statement

Let $G$ be a finite simple graph. If $M$ is a union of vertex sets of connected components of $G$ , then $M$ is a module of $G$ , and every vertex outside $M$ is anticomplete to $M$ . If $M$ is a union of anticomponents of $G$ , then $M$ is a module of $G$ , and every vertex outside $M$ is complete to $M$ .

Facts & Assumptions

Given: A finite simple graph $G$ .

[F1]

$M$ is a module of $G$ when the pair $({v}, M)$ is pure for every $v \in V (G) ∖ M$ (Modules of a graph, and the trivial modules).

[L1]

The vertex sets $C_{G} (v)$ of the connected components of $G$ are nonempty, cover $V (G)$ , and any two are equal or disjoint (The connected components of a graph partition its vertex set and are its maximal connected subgraphs, Connected graphs and connected components defined by the existence of vertex paths).

[L2]

Distinct connected components of a graph are anticomplete to one another, and distinct anticomponents are complete to one another (Distinct connected components are anticomplete, and distinct anticonnected components are complete).

[F2]

The anticomponents of $G$ are exactly the connected components of $\overline{G}$ ; consequently their vertex sets are nonempty, cover $V (G)$ , and any two are equal or disjoint (The anticonnected components of $G$ are exactly the connected components of $\overline{G}$ , The connected components of a graph partition its vertex set and are its maximal connected subgraphs).

[F3]

The pair $(A, B)$ of disjoint sets is complete when every vertex of $A$ is adjacent to every vertex of $B$ , anticomplete when no vertex of $A$ is adjacent to any vertex of $B$ , and pure when it is complete or anticomplete (Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs).

Proof

technique · direct

1.1

Let $M$ be a union of component vertex sets and let $v \in V (G) ∖ M$ . The component $C_{G} (v)$ meets no component contained in $M$ , since $v \notin M$ and distinct components are disjoint, so $C_{G} (v)$ is distinct from every component inside $M$ .

1.2

Let $M$ be a union of anticomponents and let $v \in V (G) ∖ M$ . The anticomponent containing $v$ is disjoint from every anticomponent inside $M$ and hence distinct from each of them.

2.1

In the component case of step 1.1, $v$ has no neighbour in any component inside $M$ , so $v$ has no neighbour in $M$ and $({v}, M)$ is anticomplete, hence pure.

step 1.1L2F3

2.2

In the anticomponent case of step 1.2, $v$ is adjacent to every vertex of every anticomponent inside $M$ , so $v$ is adjacent to every vertex of $M$ and $({v}, M)$ is complete, hence pure.

step 1.2L2F3

3.1

Steps 2.1 and 2.2 verify the condition of [F1] at every vertex outside $M$ in the two cases, so both kinds of union are modules, with the stated purity.

step 2.1step 2.2F1∎

Depends on

Used by

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Habib and C. Paul, A Survey on Algorithmic Aspects of Modular Decomposition, sec. 2.3 (standard reference, not scraped)